question_answer

When \[{{U}^{238}}\] nucleus originally at rest, decays by emitting an alpha particle having a speed u, the recoil speed of the residual nucleus is     AIEEE  Solved  Paper-2003

A) \[\frac{4\,u}{238}\]                                        

B) \[-\frac{4\,u}{238}\]                      

C) \[\frac{4\,u}{234}\]        

D)       \[-\frac{4\,u}{238}\]

Correct Answer: C

Solution :

                                   Apply conservation of linear momentum, we get \[0=4u-234v\] (-ve sign is for opposite direction) \[\Rightarrow \,\,v=\frac{4u}{234}\] The residual nucleus will recoil with a velocity of \[\frac{4u}{234}\] unit. Recoil speed of residual nucleus is \[\frac{4u}{234}\]. NOTE If they ask the recoil velocity, then answer remains same i. e., \[\frac{4u}{234}\] and not \[-\frac{4u}{234}\] as the word recoil itself is signifying the direction of motion of residual nucleus.